Proof of $n \int_0^1 f(x+t) g(nt)\, dt \rightarrow f(x)$ Suppose $g\in L^1([0,\infty))$ and $\int_0^\infty g(x)\, dx=1$. 
How to prove that if $f:[0,\infty) \rightarrow \mathbb{R}$ is a continuous function then $n \int_0^1 f(x+t) g(nt)\, dt \rightarrow f(x)$ as $n\rightarrow \infty$ for $x\in \mathbb{R}$? 
Say $nt=u$ in the integral then we get $\int_0^n f(x+u/n) g(u) \,du$. Estimating this integral from above and letting $f_n(x)= f(x+u/n) \chi_{[0,n]} $ gives nothing since we don't know whether $f$, being pointwise limit of this sequence, is integrable on $\mathbb{R}$. How can I use the continuity of $f$? Somehow Lebesgue Dominated Convergence Theorem is gonna play a role but I cannot get it, could you help? 
Obrigado.
 A: I can't help but see these questions in terms of probability.
Let $T$ be a non-negative random variable with density $g(t)$;
 then $ng(nt)$ is the density of $T/n$, and so 
$$n\int_0^1f(x+t)g(nt)\,dt=\int f(x+T/n)\,1_{[0,1]}(T/n)\,dP.$$
The random variables $f(x+T/n)\,1_{[0,1]}(T/n)$ are uniformly 
bounded by $M:=\sup_{x\leq y\leq x+1}|f(y)|$ and converge to 
the constant $f(x)$ so by dominated convergence we have
$$ \int f(x+T/n)\,1_{[0,1]}(T/n)\,dP\to \int f(x)\,dP=f(x).$$
A: This is just Byron's answer again:
Let $g_n(t) = f(x+t/n)\chi_{[0,n]}g(t)$. Note that for $t\in[0,n]$ we have  $ x+t/n\in[x,x+1]$. Let $M$ be an upper bound for $|f|$ on $[x,x+1]$. We then have $|g_n(t)|\le M |g(t)|$ for all $t\ge0$. Also $\lim\limits_{n\rightarrow\infty} g_n(t)= f(x)g(t)$   for all $t\ge0$. 
Thus, by the Dominated Convergence Theorem:
$$
\int_0^n f(x+t/n) g(t)\,dt=\int_0^\infty g_n(t)\,dt\ \ \buildrel{n\rightarrow\infty} \over\longrightarrow\ \ \int_0^\infty f(x)g(t)\,dt=f(x).
$$
A: You can easily prove the statement for linear combinations of indicator functions. Take $g=\frac{1}{b-a}\chi_{[b-a]}$:
$$
n\int_0^1 f(x+t)g(nt)dt=\frac{n}{b-a}\int_{a/n}^{b/n}f(x+t)dt=f(x+c),
$$
for a $c\in[a/n,b/n]$ (this is the meanvalue theorem). For $n\rightarrow\infty$ $c$ converges to $0$ and so $f(x+c)\rightarrow f(x)$. Now approximate $g$ by such functions to get the result.
